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2.5 piece of lithium is dropped into a 500 g sample of water at 22.1 C.ten temperature of the water increases to 23.5 C. How much heat energy is released in the reaction

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Answer:

2.6 kJ  

Explanation:

The formula for the amount of heat (q) absorbed by the water is

q = mCΔT

1. Calculate ΔT

ΔT = 23.5 °C - 22.1 °C = 1.4 °C

2. Calculate q

q₂ = mCΔT = 500 g × 4.184 J·°C⁻¹g⁻¹ × 1.4 °C = 2900 J = 2.9 kJ

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